JEE Main Solved Paper 2017 Question 24
Question: If for $ x\in ( 0,\frac{1}{4} ), $ the derivative of $ {{\tan }^{-1}}( \frac{6x\sqrt{x}}{1-9x^{3}} ) $ is $ \sqrt{x}.g(x), $ then $ g(x) $ equals:- [JEE Main Solved Paper-2017]
Options:
A) $ \frac{3}{1+9x^{3}} $
B) $ \frac{9}{1+9x^{3}} $
C) $ \frac{3x\sqrt{x}}{1-9x^{3}} $
D) $ \frac{3x}{1-9x^{3}} $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ y={{\tan }^{-1}}( \frac{6x\sqrt{x}}{1-9x^{3}} ) $ where $ x\in ( 0,\frac{1}{4} ) $
$ ={{\tan }^{-1}}( \frac{2.(3{x^{3/2}})}{1-(3{x^{3/2}})} )=2{{\tan }^{-1}}(3{x^{3/2}}) $
As $ 3{x^{3/2}}\in ( 0,\frac{3}{8} ) $
$ \therefore $ $ g(x)=\frac{9}{1+9x^{3}} $
